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I have a set of relations, shown below:

$R=\{(a,b), (b,c), (c,a)\}$ for $A= \{a,b,c\}$

According to my professor, this relation is transitive but I don't understand why. I was under the impression that any given three relations of a set of relations have to follow the $aRb, bRc, aRc$ pattern but since $aRc$ isn't in the set then it'd fail the test since $aRc$ and $cRa$ aren't equivalent. I'm sure I'm missing something, but I don't understand what.

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    $\begingroup$ Probably there is a typo and $(c,a)$ should be $(a,c)$. The relation as written is not transitive. In fact, there are three reasons: (1) $aRb\wedge bRc$ but not $aRc$; (2) $bRc \wedge cRa$ but not $bRa$; (3) $cRa\wedge aRb$ but not $cRb$. $\endgroup$ – vadim123 May 27 '13 at 22:13
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    $\begingroup$ Either there's a typo, as Vadim pointed, or your prof. is wrong. $\endgroup$ – DonAntonio May 27 '13 at 22:17
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    $\begingroup$ You don't have a set of relations, you have only one relation, $R$. $\endgroup$ – Prateek Dec 30 '13 at 14:56
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As expressed, the relation $R$ fails to be transitive. Indeed, we need $(a, c) \in R$ and instead of $(c, a) \in R$, as $(c, a)$ poses a number of problems in terms of the relation being transitive:

As written:

  • $(a, b), (b, c) \in R,$ but $(a, c) \notin R$, as you observe.
  • Furthermore: $(c, a), (a, b) \in R$ but $(c, b) \notin R$.
  • And, $(b, c), (c, a) \in R$, but $(b, a)\notin R$.

For each reason above, transitivity fails.

So $(c, a) \in R$ seems to be the ill-placed, perhaps misprinted pair in $R$, which if replaced by $(a, c)$ would alleviate these failures, and the relation would then be transitive.

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  • $\begingroup$ Thanks for the quick responses, everybody. I noticed this practice exam that we received in class has solutions with a few (assumed) typographical issues. A follow up question: Is a set of relations R={(a,a), (b,b), (c,c)} also not transitive? At a glance I want to say no since I can't see any logical link that would let me say aRc, but I'm really not very good at mathematics as a whole and this is all one big uphill battle for me $\endgroup$ – Harry Tuttle May 27 '13 at 23:00
  • $\begingroup$ Yes, to your follow up question: it is vacuously true that it is transitive, since there are no counterexamples for which it is NOT transitive. A relation has a property, unless a counterexample exists. Your follow-up relation is symmetric, for the same reason. $\endgroup$ – Namaste May 27 '13 at 23:23
  • $\begingroup$ Excellent information. Thank you! $\endgroup$ – Harry Tuttle May 27 '13 at 23:59
  • $\begingroup$ You're welcome, Harry! Just hang in there...you'll get the hang of the class! $\endgroup$ – Namaste May 28 '13 at 0:00
  • $\begingroup$ @amWhy: nice discourse ... +1 $\endgroup$ – Amzoti May 28 '13 at 0:36

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